Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




Not registered yet?

#26 2006-08-23 02:18:17

Full Member


Re: Lie groups anyone?

Oh? Then you are lot smarter than me. Again, I'm working till 11pm UK, but I really want to say a bit more about homeomorphism before moving on. But first assure me that you understand this generalization of the usual δ/ε defintion of continuity of a function.

Let X, Y be topological spaces.  If f: X → Y, then f is continuous at x if for every open set ε containing y = f(x), there is an open set δ containg x such  for every ε in Y, f-(ε) is open in X (f-(ε) is the preimage in X of ε in Y),  Gotta run now. Oh remember, "open set containing x" = "neighbourhood of x"

Last edited by ben (2006-08-23 02:43:44)


#27 2006-08-23 18:19:28



Re: Lie groups anyone?

Thought I'd throw in my 2p.  Where topological spaces come from and how there are really two ways of saying a set is open. Sorry if you already know all this.

The usual definition of an open set in R (before thinking about topological spaces) is:
A is open if for all x in A we can find ε>0 such that {y in R : |x-y|<ε} is a subset of A. 
So given any point in S (no matter how close to the edge) there's always room to move around without leaving S.  Obviously if really close to the edge it won't be much room but it'll always exist.

The next step is to ask what an open set in something other than R is.  We used |x-y| in the definition of open but that won't work in general as | | only applies to R (and R etc).  That's where metric spaces come in.  All we really need is a notion of "distance between two points" to be able to say a set is open. 
A metric space is a set S together with a function d:SxS→R satisfying:
d(x,y)≥0 for all x,y in S
d(x,y)=0 iff x=y
d(x,y)=d(y,x) for all x,y in S
d(x,z)≤d(x,y)+d(y,z) for all x,y,z in S (the triangle inequality)
Basically, it satisfies all the stuff you'd expect a "distance" function to satisfy.
The metric space would be written (S,d). 
See that d(x,y) = |x-y| defines a metric on R.

So now we can define "open set" in general.  If A is a subset of (S,d) then it's open if for all x in A we can find ε>0 such that {y in S : d(x,y)<ε} is a subset of A

We can also talk about continuity:
Let (A,a) and (B,b) be metric spaces.  f:A→B is continuous if for all x in A and for all ε>0 there exists δ>0 such that y is in A and d(x,y)<δ => d(f(x),f(y))<ε

One thing to note is that there can be different metrics on A and B and f may be continuous with respect to one pair of metrics but discontinuous with respect to another.  Similarly, a set may be open in one metric and not open in another.  So you have to be careful.  For example.
Define a metric on R by d(x,y) = 0 if x=y and 1 if not.  Check this is a metric (called the discrete metric). What are the open sets? Let A be a subset of R. For any x in A take ε=.  Then {y in R : d(x,y)<ε} = {x} which is clearly a subset of A.  So by definition A is open.  Since A is ANY subset of R we have proved that under this metric any set is open in R.  This is clearly not true under the usual modulus metric.

Open sets still satisfy the properties they do in R. ie:
If A and B are open sets then their intersection is open. (Finite intersection is still open)
The union of any arbitrary collection of open sets is open.  (This allows for the union of an infinite number of open sets)
The empty set and the whole space are open.

Next step is to throw away the metric.  The idea is that the metric isn't important.  It's the behaviour of open sets under intersection and union (above) that's important.  This is how a topological space is defined.  You simply decide which subsets you want to call open and as long as they satisfy the above you can talk about continuous maps and compactness and conectedness or whatever, just as you could in a metric space.

Continuity can be stated in terms of the open sets alone:
Let A and B be topological spaces.  f:A→B is continuous iff whenever U is open in B, f-(U) is open in A. 
Applying this definition to metric spaces you can show that it agrees with the usual ε-δ definition.  Topological spaces are just an abstraction of that.  Again, this depends on the topologies.  May be continuous with one pair and not with another.

There was some confusion over the "standard" topology on R.  We know what an open set in R looks like.  If we take our topology to be the collection (that is collection, not union) of usual open sets we get the standard topology on R.  I think the problem came when deciding whether [1,2) was open or closed or neither or both. 
"It's neither because it's not one of the sets in the topology".  "So the topology is defined by taking the open sets and the open sets defined by the topology... kinda circular".  Something like that.  Well this is where two senses of the word open come into play.  The standard topology is defined by taking the open sets and the open sets are defined in the usual way (with the modulus).  You could define a different topology with the sets looking like [a,b).  Then the set [m,n) would be open in the topology but neither open or closed in the usual sense.
When talking about general topological spaces the phrase "U is open" just means "U is one of the sets in the collection that makes the topology" or "U is one of the sets in the topology".

Oh.  If A is a subset of T then A is closed in T if its complement T\A is open.

And remember the discrete metric.  And how all subsets are open under it.  If S is a set and the collection of subsets in our topology is ALL the subsets of S (all subsets are "open") it's called the discrete topology on S.  Mentioned in the first post. 

We can define a topology from a metric by taking our collection of subsets to be those which are open under the metric.  The metric is said to "induce" the topology on the set.  On the other hand, if we are given a topological space and we can find a metric that would induce it, that topology is said to be "metrizable".

Um.  I think I'll go now.  Hope this helped someone. smile

Last edited by SsEe (2006-08-23 18:50:18)


#28 2006-08-23 20:41:35

Full Member


Re: Lie groups anyone?

ben wrote:

Let X, Y be topological spaces.  If f: X → Y, then f is continuous at x if for every open set ε containing y = f(x), there is an open set δ containg x such  for every ε in Y, f-(ε) is open in X (f-(ε) is the preimage in X of ε in Y),

In my rush to get off last night, I forgot to add the following qualifier......"and δ subset f-(ε), or f(δ) subset ε".

SsEe Excellent post. I am aware that it is common to start with a metric space and then chuck out the metric. It is not the approach I favour, I prefer to work by analogy rather than deconstruction, but it's merely a personal preference.

Not all spaces are metrizable, of course. In fact the ones I wanted to taalk about are not. A couple of points, however.

SsEe wrote:

There was some confusion over the "standard" topology on R.  We know what an open set in R looks like.  If we take our topology to be the collection (that is collection, not union) of usual open sets we get the standard topology on R

I think you misread my post. I had said that open sets in the standard topology on R is all unions of open intervals in R, not the union of open sets. (Hmm, looking back, I see I didn't quite say that, but nearly!)

If A is a subset of T then A is closed in T if its complement T\A is open.

This is misleading, as I used T for the topology on the set S. Under this nomenclature, all sets in T are open. Are you using T for the whole space? Let (S,T) be a topological space. A subset A of (S,T) is closed in (S,T) if its complement (S,T) - A is open i.e. in T.

Last edited by ben (2006-08-23 20:46:17)


#29 2006-08-24 07:35:31



Re: Lie groups anyone?

Ok.  So you are saying the open intervals are a basis for the standard topology on R.  Which is exactly the same as saying the topology is the collection of open sets in R since all open sets in R can be expressed as the union of open intervals.

Sorry.  Inconsistency in my notation.  I meant T to be the whole space.  What you said is better.


#30 2006-08-24 08:24:57

Full Member


Re: Lie groups anyone?

SsEe I was trying to take things sowly, for the benefit of some very talented members who are less accustoumed to abstact algebra. But please feel free to correct my errors

Anyway, Ricky (and all others who might be interested, of course) you invited me to proceed.

I said a while back that the notion of homeomorphism is a central theme
in topology. Now seems an appropriate time to revisit this.

Let X and Y be sets, and let f: X → Y. Recall that f is a surjection
(an onto map) if, for every y in Y there is at least one x
in X such that f(x) = y. Also that f is an injection (a one-to-one map)
if, for any y = f(x) there is at most one x in X for which
f-(y) = x. If f is both surjective and injective, it is referred to as
a bijection, and we may assume there is some g: Y → X such that g(f(x))
= x for all x in X. g is thus an inverse of f, and one writes (g f): X → X.

Now using the definition of continuity of a function between
topological spaces that we agreed, if both f and g are continuous inverses of each other, one says that they are homeomorphisms, and that X and Y are

Effectively (remebering our new definition of continuity) this means that we are entitled to think of each neighbourhhod of X as equivalent to some neighbourhood of Y.
And as the neighbourhoods of x in X and y in Y are the open sets in their
respective topologies, it follows that there is a 1 to 1 correspodence
between these topologies: X and Y are thus topologically equivalent.

Topologically equivalent spaces share what are known as topological
properties. We have only mentioned the Hausdorff and conectedness
properties, but there are others, as SsEe hinted at.

And if we say that any closed, connected and Hausdorff n-dimensional
surface is topologically equivalent to any other closed, connected and
Hausdorff n-dimensional surface, we are simply showing off, and really
mean that any circle is homeomorphic to any triangle and any rectangle!
We can think of this as meaning that each can be changed into the other
without cutting or gluing. When (if!) we get to talk about manifolds,
this will be an incredibly important notion.


#31 2006-08-27 04:10:47

Full Member


Re: Lie groups anyone?

Ok, let's get to the heart of this. I'm not going to do any math today, just to orientate you a little. Well...a manifold is a topological space which can be covered by coordinate neighbourhoods. That's enough of that.

So. Ever used a road map? Do you think that Earth is flat? Well, your road map certainly is, so what's going on?

First consider the problem of trying to represent an entire 2-sphere on a plane surface. We are familiar with such maps, and we see, for example, in most there is is an infinite distance between Alaska and Siberia (is this a good thing?). Similarly, we lose information at the poles. We have the additional (and potentially more serious) problem that distance and angle cannot easily be transferred from the sphere to the plane.

And yet, we are happy to navigate using our road map. What's going on? Well, all the map makers have done is broken our sphere (Earth) into small patches where the difference between a curved surface and a flat one is ignorably samll. Some of you smarties out there will immediately think calculus - and you would be correct!

All we need do now is find a sensible way of relating a map of one small patch to that of a neighbouring small patch in such a way that we don't lose sight of the fact that our Earth is really a 2-sphere.

If there exists a toplogical space where this an be done, then (loosely)  we'll say it is a manifold.


#32 2006-08-29 06:18:38

Full Member


Re: Lie groups anyone?

In case you thought my babbling about road maps was mere analogy, let me try and show that it is much stronger than that.

I remind you that a neighbourhood of a topological space M is an open
set U containing some point p in M.

We define a homeomorphism f: U → Rⁿ. This simply says that, for each point p we can find a neighbourhood sufficiemtly small that it is indistinguishable from some Euclidean space. Think of your road map here.

Note that, topological spaces, being sets of a sort, don't have
dimensions. But locally, M "inherits" the dimension n from from Rⁿ.

Nice thing about Euclidean space is we can throw down coordinates.
Let's do that here. Then we can call the pair (U, f) a chart of
dimension n, with U now veiwed as the coordinate neighbourhood of p.

Note that f(p) is now a collection of real numbers (it's called an
n-tuple, if you care) which are simply the coordinate components of p.
And if M can be completely covered by a collection of such charts, called
appropriately enough, an atlas, we will say that M is a topological
manifold (or manifold for short).

Now I said "covered". It should be easy enough to see that this implies
some notion of overlap between adjacent charts, and we will stipulate
then that each point p, q,..... in M lies in at least one
coordinate neighbourhood. (Recall here that most road maps overlap at
their edges). We see that this gives us a connected space, and let's
stipulate the Hausdorff property, just to be complete.

Note that each chart has its own coordinate set, locally deterimed by
the topology on M. Under these circumstances, consider the point p
which lies in two overlapping charts U and V. We see that p can be
described by two different coordinate systems, those on U and V, so it
must be possible to relate these to each other.
Let's try that now.....

On second thoughts, let's not. I guess that's enough to be going on with. Later, if anyone wants.


#33 2006-09-05 05:32:25

Full Member


Re: Lie groups anyone?

I am aware that I am talking to myself, but this is common. So I will
continue down this road in the way that makes sense to me.....So.

Recall my definition of a manifold. Let M be a manifold, with p a point in M and U and V be neighbourhoodsof p. Let the point p be in W = U ∩ V, and let f: U → Rⁿ and g: V → Rⁿ be homeomorphisms. Then (U, f) and (V, g) are charts on M.

Let {x, x,.....,xⁿ} = {a} be a coordinate set on the chart (U, f) and
{y, y,....,yⁿ} = {b} be a coordinate set on (V, g). How do we relate
these guys, as we must, since p lives in both. (By the way, the indices
are not exponents, just tags)

Now f and g are homeomorphisms, that is, they have continuous inverses.
So f-: Rⁿ → U and g: V → Rⁿ. Then the composite (g f-): Rⁿ → Rⁿ,
that is, we now have a way of "comparing" {a} and {b}, the two
different coordinate systems which apply equally to our point p.

Now, if we consider what (V,g) will "look like" from (U, f) we see that
each component of {b} will a function h of each component of {a}, like

y = h(x, x,....,xⁿ)
y = h(x, x,....,xⁿ)
yⁿ = hⁿ(x, x,....,xⁿ)

Now obviously, the value of each h is completely dependent on the
numerical value of the corresponding y, so we may as well have

y = y(x, x,....,xⁿ)
y = y(x, x,....,xⁿ)
yⁿ = yⁿ(x, x,....,xⁿ)

And going the other way, that is, what (U, f) looks like from (V, g) we
simply substitute x for y, and therefore see that the coordinates on U
and V are functions on the point p.

This is an important, but not very profound,  result. It gives you the whole flavour of manifolds, which are hugely important.

Anyway, I'm a bit bit tired of talking to myself, so I'm going to quit, unless I can think of something that will interest us all equally.

Until then, xxx


Board footer

Powered by FluxBB